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ValentinkaMS [17]
3 years ago
7

Please help! I need answers in begging !

Mathematics
1 answer:
Levart [38]3 years ago
7 0

Answer:

2866

Step-by-step explanation:

10000-7134=2866

so 7134+2866=10000

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Hello can anyone help me please
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5.3 to the 4th power :)

Step-by-step explanation:

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Jess deposited $5,000 into an account that earns simple interest. After 9 years, Jess had earned $3,150 in interest. What was th
musickatia [10]

Answer:

The interest rate of Jess's account was 7%

Step-by-step explanation:

A = P * (1 + rt)

A = final amount

P = initial principal balance

r = annual interest rate

t = time (in years)

Replacing with the values we know:

A =  P * (1 + rt)

8,150 = 5,000 * (1 + r * 9)

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1.63 = 1 + 9r

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The interest rate of Jess's account was 7%

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3 years ago
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What number does x stand for in this equation?
Alex777 [14]

Answer:

C. -3

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

Step-by-step explanation:

<u>Step 1: Define</u>

-7x + 6 = 27

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. [Subtraction Property of Equality] Subtract 6 on both sides:                        -7x = 21
  2. [Division Property of Equality] Divide -7 on both sides:                                 x = -3
3 0
2 years ago
A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears
marta [7]

Answer:

(a) The probability that <em>Y</em>₂₀ exceeds 1000  is 3.91 × 10⁻⁶.

(b) <em>n</em> = 28.09

Step-by-step explanation:

The random variable <em>Y</em>ₙ is defined as the total numbers of dollars paid in <em>n</em> years.

It is provided that <em>Y</em>ₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of <em>Y</em>ₙ are:

\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}

(a)

For <em>n</em> = 20 the mean and standard deviation of <em>Y</em>₂₀ are:

\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\

Compute the probability that <em>Y</em>₂₀ exceeds 1000 as follows:

P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z>  4.47)\\=1-P(Z

**Use a <em>z </em>table for probability.

Thus, the probability that <em>Y</em>₂₀ exceeds 1000  is 3.91 × 10⁻⁶.

(b)

It is provided that P (<em>Y</em>ₙ > 1000) > 0.99.

P(Y_{n}>1000)=0.99\\1-P(Y_{n}

The value of <em>z</em> for which P (Z < z) = 0.01 is 2.33.

Compute the value of <em>n</em> as follows:

z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n}  \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0

The last equation is a quadratic equation.

The roots of a quadratic equation are:

n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of <em>n</em> = 28.09.

8 0
3 years ago
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